document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=231630>Question 231630</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three married couples arrange themselves randomly in six consecutive seats in a row. Find the probability of the event of the women being in three adjacent seats, as well as the men. Hint: In this case the denominator of the probability fraction will be 6!=720 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #171395 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=171395\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Three married couples arrange themselves randomly in six consecutive seats in a row. Find the probability of the event of the women being in three adjacent seats, as well as the men. Hint: In this case the denominator of the probability fraction will be 6!=720<BR>\n" );
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document.write( "# of ways to succeed<BR>\n" );
document.write( "Note: there are 2 sets of 3 consecutive seats in a row of 6 seats<BR>\n" );
document.write( "The women can be arranged in 3! ways in their three seats<BR>\n" );
document.write( "The men can be arranged in 3! ways in their three seats.<BR>\n" );
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document.write( "# of ways to succeed = 2*3!*3! = 2*6*6 = 72<BR>\n" );
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document.write( "Probability = 72/6! = 72/720<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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